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3 years ago

How many moles would be equal to 22.4 g of CO2?

Chemistry

1 answer:

Rzqust [24]3 years ago

4 0

Answer:

Number of moles = 0.51 mol

Explanation:

Given data:

Mass of CO₂ = 22.4 g

Number of moles = ?

Solution:

Number of moles of CO₂:

Number of moles = mass/molar mass

Molar mass of CO₂ = 44 g/mol

by putting values,

Number of moles = 22.4 g/ 44 g/mol

Number of moles = 0.51 mol

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Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

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3 years ago

A chemist takes 50-gram sample of sulfur powder that has a melting point of 115.2 °C. What is the melting point of a 100-gram sa

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Identify the reactants and the products of the preparation step to the urea cycle.

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Which phrase correctly describes the outer shell of electrons of each atom in most covalent compounds?

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How many grams of FeCo3 will be produced from 57.2g FeCl2

Evgesh-ka [11]

Answer:

287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 2 mol.L⁻¹ × 1.24 L

Moles = 2.48 mol

Also,

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting Values,

Mass = 2.48 mol × 126.75 g.mol⁻¹

Mass = 314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃

So,

314.34 g of FeCl₂ will produce = X g of FeCO₃

Solving for X,

X = (314.34 g × 115.85 g) ÷ 126.75 g

X = 287.30 g of FeCO₃

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2 years ago